and many more benefits!

Find us on Facebook

GMAT Club Timer Informer

Hi GMATClubber!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

Show Tags

19 Sep 2015, 14:36

2

This post receivedKUDOS

Bunuel wrote:

If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5B. 6C. 7D. 8E. 9

Kudos for a correct solution.

Since the sequence is in a evenly spaced set,the average of the sequence of consecutive integer formula is mean=\(\frac{First Term +Last Term}{2}\) ,or 153=\(\frac{First Term+270}{2}\) ,or First Term=36

So the consecutive numbers less than mean of the sequence are 36,54,72,90,108,126,&144.So the total number if integer below the mean of the sequence is 7

Show Tags

This question can be approached in a number of different ways. Since we're given so many facts to work with, you can actually use some math 'logic', a bit of arithmetic and some 'brute force' to PROVE how many terms are below the average.

First, we're given a series of facts:1) We're dealing with CONSECUTIVE multiples of 182) The average of all those multiples of 18 is 1533) The LARGEST of those multiples is 270

We're asked to for the number of terms in that sequence that are BELOW the average.

To start, since we're dealing with consecutive terms, the SAME number of terms will be above the average as below the average. Since the largest term is 270, and the average is 153, we can determine the number of terms that are ABOVE the average (we just have to keep subtracting 18 to find all of those terms....):

270252234216198180162

At this point, we can stop. If we subtract another 18, we'll end up with 144 - which is BELOW the average. Since we have 7 terms that are ABOVE the average, and we're dealing with CONSECUTIVE multiples of 18, there has to be the SAME NUMBER of terms below the average (to 'balance out' the calculation).

Show Tags

If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms in the sequence are smaller than the average?

A. 5B. 6C. 7D. 8E. 9

Kudos for a correct solution.

given that sequence of consecutive multiple of 18 and avg is 153 therefore median = 153 greatest 270

therefore visualize this X------- 153--------270 Where x is min

and 270 is 15th term of the sequence as odd number so 8th term will the median ---- and therefore there are 7 terms less than the median

ans is c

Hi,abhisheknandy08,the coloured portion is not correct..you cannot say that 270 is the 15th term since you do not know the first term....you have to find the first number through average and last number given...
_________________

Re: If the average of a sequence of consecutive multiple of 18 is 153, and [#permalink]

Show Tags

25 Jun 2016, 23:29

In series of consecutive number avg is is defined as avg of two middle terms , if numbers of terms are evenand avg is defined as middle term , if number of terms are odd.153 is odd number , which cannot be the multiple of 18 , which is even number; hence this has to be the avg of middle even terms.

Number of terms above is equal to number of term below the avgNext multiple of 18 is 162 last multiple given is 270270-162/18 +1 = number of terms.